Fluid dynamics •  FCQ’s for lecture will be on

Wednesday. •  Finish up fluids today •  Review on Wednesday and Friday. •  Final exam at 7:30am on Tuesday,

May 5 in Coors Event Center •  The 3 midterm exams without

answers are posted on D2L. Hard copies of the 3rd exam are available in the front.

Some mercury will run out, creating a vacuum in the top.

Mercury barometer Flip a tube filled with mercury upside down in a bucket of mercury.

Let’s examine the pressure at the dotted line.

Outside the tube, pressure is atmospheric: p = p0

Inside the tube, pressure is p = 0+ ρgh

h

These two are at the same height so must be equal: p0 = ρgh

Atmospheric pressure holds up the column of mercury

The height of the column measures atmospheric pressure

1 atm = 105 Pa = 760 mmHg = 760 torr = 30 inHg = 14.7 psi

Atmospheric pressure pushes! The vacuum does not suck!

What is the force on the top due to the atmosphere?

What happens when the air is evacuated from a 55 gallon drum?

I will use English units for this exercise.

Atmospheric pressure is 14.7 psi = 14.7 lb/in2. h = 34 in

in 12=r

F = pA = pπr2 =14.7 lb/in2 ⋅3.14 ⋅ 12 in( )2= 6600 pounds

This is the weight of an H2 Hummer!

Usually have the same air pressure underneath the top so the net force is 0. Applies to the sides and bottom as well.

Clicker question 1 Set frequency to BA

A. Nothing B. Squashed like stepping on a soda can C. Implodes from all sides D. Explodes

What happens when that air pressure is removed?

4

Fluid flow So far, all of our fluid studies have dealt with static situations (the fluid was in static equilibrium).

Today we investigate flowing fluids.

We only consider ideal fluids. Ideal fluids…

are incompressible – good assumption for liquids, not for gases.

are nonviscous – no kinetic friction between liquid parts

have laminar flow – steady flow with no turbulence

5

Streamlines Consider water flowing through a tube.

One can imagine paths taken by a particle in the fluid. These paths are streamlines.

In an ideal fluid, streamlines have two important properties:

1. Streamlines never cross.

2. Fluid speed is higher where streamlines are closer together.

This is a continuity equation which says whatever goes in must come out; fluid is not created or lost inside.

Continuity equation For a steady flow, the amount of fluid entering point 1 in time Δt must be equal to the amount of fluid exiting point 2 in time Δt.

How much fluid goes through points 1 and 2 in time Δt?

An area A traveling at a speed of v will travel d = vΔt in time Δt.

1 2

A1

d1 = v1Δt

A2

d2 = v2Δt For steady flow, the two volumes A1d1 & A2d2 must be equal: tvAtvA Δ=Δ 2211

Therefore: A1v1 = A2v2

A hose has a diameter of 4 cm. If one wants the velocity of the water coming out to be 4 times higher, what diameter nozzle should be placed on the end of the hose?

Since area depends on the diameter squared, the diameter needs to be reduced by a factor of 2 to get a factor of 4 reduction in area.

Clicker question 2 Set frequency to BA

A. ¼ cm B. ½ cm C. 1 cm D. 2 cm E. 4 cm

Solving the continuity equation for the unknown final area gives us:

A1v1 = A2v2A2 =

A1v1v2

If we want v2/v1 = 4 then we need A2 = A1/4.

In equations: A2 =A1

4 becomes πd2

2

4=πd1

2

4 ⋅ 4 or d2

2 =d1

2

4 so d2 =

d1

2= 2 cm

8

About the continuity equation The continuity equation says that the same amount (volume) of liquid goes in as comes out. Note that area times velocity (Av) has units of volume/second.

Thus, knowing this could tell us how quickly a vessel of a given volume may be filled or emptied.

Continuity equations show up where there is a conserved current. Here it is liquid but next semester it will be electrical current.

9

The water entering feels a force from the pressure of p1A1 over a distance of d1 so the work done is W=p1A1d1=p1V

Bernoulli’s equation

We consider each term with regards to fluid flow starting with Wext.

Consider two equal volumes V of water entering and leaving a tube.

p1 p2 A1 A2

d1 d2

The water exiting feels a force of p2A2 over a distance of d2 but force and displacement are in opposite directions so the work done is negative: W = –p2A2d2 = –p2V

Wext = ΔK +ΔUOur general conservation of energy equation looks like K1 +U1 +Wext = K2 +U2 or

The net work done by pressure is W = p1V – p2V

10

Bernoulli’s equation

y1

y2

A1

A2

d1

d2

Now that we have all three components, we can combine them.

We are dealing with the same volumes of water: V = A1d1 = A2d2.

Gravitational potential energy is mgy.

For a volume V of liquid m = ρV

So at points 1 and 2 we have U1 = ρVgy1 U2 = ρVgy2and

Kinetic energy is ½mv2 and m = ρV so and K1 =

12 ρVv1

2 K2 =12 ρVv2

2

11

Bernoulli’s equation

Gravitational potential energy: and U1 = ρVgy1 U2 = ρVgy2Kinetic energy: and K1 =

12 ρVv1

2 K2 =12 ρVv2

2

The net work done by pressure is Wext = p1V − p2V

Conservation of energy equation: K1 +U1 +Wext = K2 +U2

Conservation of energy: 12 ρVv1

2 + ρVgy1 + p1V − p2V = 12 ρVv2

2 + ρVgy2

Dividing through by the volume V and rearranging gives

Bernoulli’s equation: p1 +12 ρv1

2 + ρgy1 = p2 +12 ρv2

2 + ρgy2

Bernoulli’s equation is a restatement of conservation of energy!

Problems often require both Bernoulli & continuity equations.

Fluid flow equations

Rewriting conservation of energy using density times volume instead of mass and accounting for work done by the force from pressure gives Bernoulli’s equation.

p1 +12 ρv1

2 + ρgy1 = p2 +12 ρv2

2 + ρgy2

Problems often require both Bernoulli & continuity equations.

The continuity equation tells us that what goes in must come out A1v1 = A2v2

Continuity equation

Bernoulli’s equation

p+ 12 ρv

2 + ρgy = constantor

Explaining lift Wings on a plane and sails on a boat both use Bernoulli’s equation. Air travels faster over one side resulting in a smaller pressure which creates a force called lift.

Also explains effect of curve ball and ball in leaf blower.

Faster moving air leads to lower pressure which leads to less force on that side.

14

Bernoulli’s equation example m 2001 =y

m 1502 =y

03 =y

Dam 1

3

2

Consider the dam shown. The inlet pipe at point 2 has a radius of 10 m while the outlet pipe at point 3 is half that. What is the water velocity at 3 and the pressure at 2?

Use Bernoulli’s equation at points 1 & 3.

p1 +12 ρv1

2 + ρgy1 = p3 +12 ρv3

2 + ρgy3

Pressure is atmospheric at both 1 & 3. Velocity is 0 at point 1. p0 + ρgy1 = p0 +

12 ρv3

2 so v3 = 2gy1 = 63 m/sρgy1 =12 ρv3

2and thus

This is the same velocity as dropping an object from a height y1.

15

Bernoulli’s equation example m 2001 =y

m 1502 =y

03 =y

Dam 1

3

2

What is the pressure at 2?

Bernoulli’s equation at 1 & 2:

p0 + ρgy1 = p2 +12 ρv2

2 + ρgy2

Still need v2. Use continuity equation to get it.

Point 2 has r=10 m. Point 3 has r=5 m.

Solve for p2: p2 = p0 + ρgy1 −12 ρv2

2 − ρgy2

. Since radius at 2 is twice as big as at 3 and area goes as radius squared, the velocity is ¼ as great. A2v2 = A3v3

v2 = v3 / 4 = 2gy1 / 4

p2 = p0 + ρgy1 −12 ρv2

2 − ρgy2 = p0 + ρgy1 −116 ρgy1 − ρgy2

= p0 +1516 ρgy1 − ρgy2 = p0 + ρg(

1516 y1 − y2 )

=105 Pa +1000 kg/m3 ⋅10 m/s2 1516 200 m −150 m( ) = 475, 000 Pa

This is less than the hydrostatic pressure: p = p0 + ρgh = 600, 000 Pa